# Minimal Ring Generated by System of Sets

## Contents

## Theorem

Let $\SS$ be a non-empty system of sets.

Then there is a unique ring of sets $\map \RR \SS$ which:

- $(1): \quad$ contains $\SS$
- $(2): \quad$ is contained by every ring of sets which also contains $\SS$.

This ring of sets $\map \RR \SS$ is called the **minimal ring generated by $\SS$**.

## Proof

### Uniqueness

Suppose there were two such rings of sets $\map \RR \SS$ and $\map \RR \SS'$.

Then by definition their intersection $\map \RR \SS \cap \map \RR \SS'$ would also contain $\SS$.

By Intersection of Rings of Sets, $\map \RR \SS \cap \map \RR \SS'$ is also a ring of sets.

From Intersection is Subset, $\map \RR \SS \cap \map \RR \SS' \subseteq \map \RR \SS$ and $\map \RR \SS \cap \map \RR \SS' \subseteq \map \RR \SS'$.

Hence either $\map \RR \SS$ or $\map \RR \SS'$ can not be minimal.

So if $\map \RR \SS$ exists, it has to be unique.

### Existence

Consider the union:

- $\displaystyle X = \bigcap_{A \mathop \in \SS} A$

of all sets in $\SS$.

Now consider the power set $\powerset X$ of all subsets of $X$.

From Power Set is Algebra of Sets and by definition of algebra of sets, $\powerset X$ is a ring of sets containing $\SS$.

Now let $\Sigma$ be the set of all rings of sets contained in $\powerset X$ which also contain $\SS$.

Then consider the intersection:

- $\displaystyle \map \RR \SS = \bigcap_{\TT \mathop \in \Sigma} \TT$

of all these rings of sets.

It is clear that $\map \RR \SS$ has the properties specified.

In particular, as all $\TT \in \Sigma$ contain $\SS$, then so does $\map \RR \SS$.

Finally, note that if $\RR'$ is a ring of sets containing $\SS$, then:

- $\RR^* = \RR' \cap \powerset X$

is a ring of sets in $\Sigma$.

Hence $\map \RR \SS \subseteq \RR^* \subseteq \RR'$ as we needed to show.

$\blacksquare$